a very occasional diary: a trivial exercise.

Let's find a sum

There is a well-know standard way, that I managed to recall eventually. Given that

${1 \over n(n+2)} = {1 \over 2}\cdot \left({1\over n} - {1\over n+2}\right)$
the sum can be re-written as

$\sum_{n=1}^\infty {1\over{n(n+2)}} = \sum_{n=1}^\infty {1 \over 2}\left({1\over n} - {1\over n+2}\right) = {1\over 2}\left({1\over 1} - {1\over 3} + {1\over 2} - {1\over 4} + {1\over 3} - {1\over 5} + {1\over 4} - {1\over 6} \cdots\right)$
with almost all terms canceling each other, leaving

$\sum_{n=1}^\infty {1\over{n(n+2)}} = {1\over 2}\left(1 + {1\over 2}\right) = {3\over 4}$
While this is easy to check, very little help is given on understanding how to arrive to the solution in the first place. Indeed, the first (and crucial) step is a rabbit pulled sans motif out of a conjurer hat. The solution, fortunately, can be found in a more systematic fashion, by a relatively generic method. Enter generating functions.

First, introduce a function

$f(t) = \sum_{n=1}^\infty {t^{n + 1}\over n}$
The series on the right converge absolutely when |t| < 1, so one can define

$g(t) = \int f(t) dt = \int \sum_{n=1}^\infty {t^{n + 1}\over n} = \sum_{n=1}^\infty \int {t^{n + 1}\over n} = \sum_{n=1}^\infty {t^{n + 2}\over {n(n+2)}} + C$
with the sum in question being

$\sum_{n=1}^\infty {1\over{n(n+2)}} = g(1) - C = g(1) - g(0)$
Definition of the g function follows immediately from the form of the original sum, and there is a limited set of operations (integration, differentiation, etc.) applicable to g to produce f.

The rest is more or less automatic. Note that

$- ln(1 - t) = t + {t^2\over 2} + {t^3\over 3} + \cdots$
so that

$f(t) = t^2 + {t^3\over 2} + {t^4\over 3} + \cdots = - t \cdot ln(1-t)$
therefore

$g(t) = - \int t \cdot ln(1-t) dt = \cdots = {1\over 4} (1 - t)^2 - {1\over 2} (1 - t)^2 ln(1 - t) + (1 - t) ln(1 - t) + t + C$
where the integral is standard. Now,

$g(1) - g(0) = 1 - {1\over 4} = {3\over 4}$
Voilà!

And just to check that things are not too far askew, a sub-exercise in a pointless programming:

scala> (1 to 10000).map(x => 1.0/(x*(x+2))).reduceLeft(_+_)
res0: Double = 0.749900014997506

PS: of course this post is an exercise in tex2img usage.

a very occasional diary.

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2009-08-21

a trivial exercise.

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